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Calculate the force between two objects that have masses of 70 kilograms and 2,000 kilograms separated by a distance of 1 meter.

Makovka662 [10]

$▪▪▪▪▪▪▪▪▪▪▪▪▪ {\huge\mathfrak{Answer}}▪▪▪▪▪▪▪▪▪▪▪▪▪▪$

The Gravitational Force between given objects will be ~

$9.34 \times {10}^{ - 6} \: \: N$

$\large \boxed{ \mathfrak{Step\:\: By\:\:Step\:\:Explanation}}$

We know that ~

$\huge\boxed{\mathrm{F = \dfrac{ Gm_1m_2}{ r²}}}$

where ~

F = gravitational force

$m_1$ = mass of 1st object = 70 kg

$m_2$ = mass of 2nd object = 2000 kg

G = gravitational constant = $6.674 × {10}^ {-11}$

r = distance between the objects = 1 m

Let's calculate the force ~

$F = \dfrac{6.674 \times 10 {}^{ - 11} \times 70 \times 2000 }{1 {}^{2} }$

$F =6.674 \times 7 \times 2 \times 10 { }^{ - 11} \times 10 {}^{4}$

$93.436 \times 10 {}^{ - 7}$

$9.3436 \times 10 {}^{ - 6} \: \: newtons$

6 0

3 years ago

Calculate the electrostatic force that a small sphere A, possessing a net charge of 2.0 x 10^-6 Coulombs exerts on another small

Anni [7]

Answer:

F = 5.33*10^-4N

Explanation:

to find the electrostatic force you use the Coulomb's law, given by the formula:

$F=k\frac{q_Aq_B}{r^2}$

k: Coulomb's constant = 8.89*10^9 Nm^2/C^2

q_a: charge of A = 2.0*10^{-6}C

q_B: charge of B = -3.0*10^{-6}C

r: distance between the spheres = 10.0m

By replacing all these values you obtain:

$F=(8.89*10^9Nm^2/C^2)\frac{(2.0*10^{-6}C)(-3.0*10^{-6}C)}{(10.0m)^2}=5.33*10^{-4}N$

hence, the forcebetween the spheres is about 5.33*10^-4N

3 0

3 years ago

An engineering intern is studying the effect of temperature on the resistance of a current carrying wire. She applies a voltage

Mars2501 [29]

Answer:

I=2.80 A

Explanation:

We Know that R =R₀(1+∝ ΔT)

R=R₀ (1+3.9*10⁻³ *(61-20))

R=R₀ (1.1599)

I=V/R=V/(R₀ (1.1599)

1.4 = V/(R₀ (1.1599) ∵ equation 1

We have to calculate I when T=-88°

R =R₀(1+∝ ΔT)

R=R₀ (1+3.9*10⁻³ *(-88-20))

R=R₀ (0.5788)

I=V/(R₀ (0.5788) ∵equation 2

Dividing equation 2 by equation 1

$\frac{I}{1.4} =\frac{1.1599}{0.5788}$

I = 2.80 A

4 0

3 years ago

Read 2 more answers

An electromagnetic wave with a peak magnetic field component of 3.20 × 10−7 T carries what average power per unit area? (μ0 = 4π

IRISSAK [1]

Answer:

Explanation:

Average power per unit area =

I avg = c B₀² / 2μ₀

= 3 X 10⁸ X (3.2 X 10⁻⁷)² / 2 X 4π × 10⁻⁷

= 1.22 X 10

12 . 2 W / m²

4 0

3 years ago

Read 2 more answers

Indica qué es una propiedad específica de la materia. Además explica por qué son útiles las propiedades específicas de la materi

Katyanochek1 [597]

Answer:

Check Explanation

Comprobar explicación

Explanation:

English Translation

Indicate what a specific property of matter is. Also explain why the specific properties of matter are useful compared to the general ones.

Solution

The specific properties of matter are properties that describes the intensive properties of the system. They are properties that do not depend on or change with the extent or size of the system. They are usually obtained by dividing the generalised properties or extensive properties by the extent or size of matter to make them independent of size/extent/Mass.

Examples of specific properties include specific heat capacity, specific volume etc. They usually have units of general units/Mass units.

The specific properties of matter are more important than the general ones because

- They help in general comparisons of the properties of different materials. They are used to rank, classify and compare properties of different materials.

- They are used in reference table/data to easily record easily accessible properties of matter. It helps to record standards that are general and independent of sizes/extents/Mass, thereby keeping the reference table/data/chart precise and concise.

- They provide us with values that are easy to memorize and remember, unlike trying to cram the different properties of different masses/sizes of matter.

In Spanish/En español

Las propiedades específicas de la materia son propiedades que describen las propiedades intensivas del sistema. Son propiedades que no dependen ni cambian con la extensión o el tamaño del sistema. Por lo general, se obtienen dividiendo las propiedades generalizadas o las propiedades extensivas por la extensión o el tamaño de la materia para hacerlas independientes del tamaño / extensión / masa.

Los ejemplos de propiedades específicas incluyen capacidad calorífica específica, volumen específico, etc. Usualmente tienen unidades de unidades generales / unidades de masa.

Las propiedades específicas de la materia son más importantes que las generales porque

- Ayudan en las comparaciones generales de las propiedades de diferentes materiales. Se utilizan para clasificar, clasificar y comparar propiedades de diferentes materiales.

- Se utilizan en la tabla / datos de referencia para registrar fácilmente propiedades de materia fácilmente accesibles. Ayuda a registrar estándares que son generales e independientes de tamaños / extensiones / masa, manteniendo así la tabla / datos / tabla de referencia precisa y concisa.

- Nos proporcionan valores que son fáciles de memorizar y recordar, a diferencia de tratar de agrupar las diferentes propiedades de diferentes masas / tamaños de materia.

Hope this Helps!!!

¡¡¡Espero que esto ayude!!!

7 0

3 years ago